quired quantity, cos c= cos a cos bi ; whence by logarithmic cal culation, the value of c may be obtained. Schol. Instead of dividing by cos b, we may multiply by sec b, since the cosine and secant are reciprocals of each other, when radius equals unity (Day's Trig., Art. 93). The last equation may then be replaced by this, cos c=cos a sec b. In general, whenever a cosine occurs as a divisor, it may be replaced by the corresponding secant as a multiplier; likewise when a sine occurs as a divisor, it may be replaced by the corresponding cosecant as a multiplier. In this way too may tangents be exchanged for cotangents, and cotangents for tangents, since they are reciprocals of each other. (2) Given a=84°, B=63°; to find C. are comp. a, comp. B, comp. C; A a comp. a=tan comp. B tan comp. C, that is, whence or cos a=cot B cot C; Now, as a=84°, cos a= 9.0192346 also, tan B=10.2928341 hence cot C 9.3120687, the logarithmic ra dius 10 being rejected: (3) Given b=37° 13′, B=43° 41'; to find C. Here comp. B is the middle part, and the other parts, b, comp. C, are opposite. Then, sin comp. B=cos b cos comp. C, cos B=cos b sin C; that is, these two angles having the same sine, because the sine of any angle is the same with that of its supplement. This is one of those cases in which two parts besides the right angle of a right-angled triangle are insufficient to determine another part. (4) Given a=113° 40', b=60° 27'; to find C. Here comp. C is the middle part; and the other parts, comp. a, b, are adjacent. Then, but, B cos C-cot a tan b: and therefore, Now as a is an obtuse angle, its cotangent is negative (Day's Trig., Art. 201), and as tan b is positive, the product of the two is negative; therefore cos C is negative, and C must accordingly be an obtuse angle. The obtuse angle answering to the number 9.8882214 in' the table of cosines is 140° 37′ 51′′, which is therefore the required angle C. (5) Given a, B; to find the other parts of the triangle. First, to find b, take b for the middle part; then comp. a, comp. B are opposite parts, and sin b sin a sin B; whence b may be calculated. Next, to find c, take comp. B for the middle part; then comp. a and c are adjacent parts; and whence cos B=cot a tan c, tan c=cos B tan a. Lastly, to find C, take comp. a for the middle part; in which case, comp. B and comp. C are adjacent parts: cos a=cot B cot C; then whence cot C=cos a tan B. If a=72° 21′ 40′′, and B=14° 10′ 25′′, in the above equations; then b=13° 29′ 40", c=71° 50' 35", and C=85° 37' 26". OBLIQUE-ANGLED TRIANGLES. 14. Of the sides and angles in an oblique-angled triangle, it is necessary, as has before been stated, that three be given, in order that the triangle may be determined. These may be (1) Two sides and the included angle, (2) Two sides and an angle opposite one of them, (3) Two angles and the intermediate side, (4) Two angles and a side opposite one of them, (5) The three angles, (6) The three sides. In the last case, every two sides are separated by an unknown angle; and in the last but one, every two angles have between them an unknown side. But in each of the other four cases, there are two at least of the given parts, which are not separated by an unknown part, but are adjacent to each other. Now every oblique-angled triangle may be divided into two right-angled triangles by a perpendicular drawn from one of the angles to the opposite side. And this may be so done as to make the two given parts that are adjacent in each of the first four of the above cases, fall in the same right-angled triangle. Then all the other parts of this right-angled triangle may be determined by Napier's rules: and thus means will be furnished for computing the unknown parts of the other right-angled triangle; whence all the parts of the oblique-angled triangle will be known. This will be seen more clearly by attending to these cases separately, as may be conveniently done after noticing the following theorem. 15. The sines of the sides of a spherical triangle are as the sines of their opposite angles. Let ABC be a spherical angle, the sine of AB is to the sine of BC as the sine of the angle ACB is to the sine of BAC. From B draw BD perpendicular to AC or AC produced, forming the right-angled triangles ABD, CBD : 1 then (Art. 5), and (Euc. 5, A), also sin BC: R:: sin BD; sin BCD, R: sin BC:: sin BCD: sin BD: sin AB:R:: sin BD: sin BAD: therefore (Euc. 5, 23), sin AB: sin BC:: sin BCD: sin BAD. Now the sines of BAC and ACB may be substituted for those of BAD and BCD, since the angles ACB and BCD, if they are not the same, are supplements of each other. Therefore sin AB sin BC:: sin ACB: sin BAC. : In like manner it may be shown that sin AC: sin AB:: sin ABC: sin ACB; wherefore also sin AC: sin BC:: sin ABC: sin BAC. 16. Given two sides and the included angle of a spherical triangle; to find the other side and angles. In the spherical triangle ABC, let the two sides AB, AC, and the angle A be given, from which to deter· mine the side BC and the angles ABC, ACB. Draw BD from B perpendicular to AC, forming two right-angled triangles ABD, BDC, such that one of them ABD has for two of its parts, two of the quantities given, namely, the side AB and the angle A. From these two, may be computed by Napier's rules, the sides AD, BD, and the angle ABD. Then DC is found by taking AD from the given side AC. And in the right-angled triangle BDC, having now given the sides BD and DC, the side BC, the angle C, and the angle CBD may be computed. The angle ABC is found by adding together the angles ABD, CBD. If the perpendicular BD meets AC produced, the method of solution is the same as has been stated, ex |